# A problem with pointwise stabilizer subgroups of fixed-point subspaces I

Definitions: Let $W$ be a representation of a group $G$, $K$ a subgroup of $G$, and $X$ a subspace of $W$.
Let the fixed-point subspace $W^{K}:=\{w \in W \ \vert \ kw=w \ , \forall k \in K \}$.
Let the pointwise stabilizer subgroup $G_{(X)}:=\{ g \in G \ \vert \ gx=x \ , \forall x \in X \}$.

Let $G$ be a finite group, $H$ a subgroup, $U$ and $V$ two irreducible complex representations of $G$.
Let the subgroup $L := G_{(U^H)} \cap G_{(V^H)}$.

Question: Is there an irreducible $W \le U \otimes V$ such that $G_{(W^H)} \cap G_{(V^H)}$ and $G_{(U^H)} \cap G_{(W^H)} \subset L$?

Let $G=C_2\times C_2$ and $H$ a subgroup of order $2$.
Let $U$ be the non-trivial irreducible module on which $H$ acts trivially, and $V$ one of the other non-trivial irreducible modules. Then $U\otimes V$ is irreducible, so $W$ can only be $U\otimes V$.
Then $G_{(U^H)}=H$, and $G_{(V^H)}=G_{(W^H)}=G$, so $G_{(W^H)}\cap G_{(V^H)}\not\subset G_{(U^H)}\cap G_{(V^H)}$.
• You're right! This formulation of the problem is too soft. In fact for my application I can assume that $H$ is a core-free subgroup of $G$, and I don't need $W \le U \otimes V$ (this assumption was just for helping, but by your counter-example it is useless because $W = U$ works whereas $W = U \otimes V$ not). See the second post A problem with pointwise stabilizer subgroups of fixed-point subspaces II. I don't think this post II admits an easy counter-example (if one exists). – Sebastien Palcoux Mar 27 '15 at 11:17